列出索引超出范围和随机数以选择列表中的项目

问题描述

我需要使用.split()从字符串中列出iPhone型号.

I need to make a list of iPhone models from a string using .split().

这不是问题，但是我还必须使用0-9之间的随机数来选择一个单词，然后使用while/for循环显示3个随机词.

That's not a problem, but I also have to use a random number from 0-9 to pick a word,then display 3 random words using while/for loop.

在我的代码中，当我输入:

In my code, when I enter:

import random

iPhone = 'Original 3G 3GS 4 4S 5 5C 5S 6 6Plus'.split()

z = 0
while z < 4:
    for y in range (1,3):
        for x in iPhone:
            x = random.randint(0,10)
            print (iPhone[x])

它说:

 Traceback (most recent call last):
      File "C:\Users\zteusa\Documents\AZ_wordList2.py", line 15, in <module>
        print (iPhone[x])
    IndexError: list index out of range

我不确定是什么原因造成的.

I'm not sure whats causing this.

推荐答案

random.randint的两个参数都包含在内:

Both arguments to random.randint are inclusive:

>>> import random
>>> random.randint(0, 1)
1
>>> random.randint(0, 1)
0
>>>

因此，当您执行x = random.randint(0,10)时，x有时可能等于10.但是您的列表iPhone仅包含十个项目，这意味着最大索引为9:

So, when you do x = random.randint(0,10), x could sometimes be equal to 10. But your list iPhone only has ten items, which means that the maximum index is 9:

>>> iPhone = 'Original 3G 3GS 4 4S 5 5C 5S 6 6Plus'.split()
>>> len(iPhone)
10
>>> iPhone[0]  # Python indexes start at 0
'Original'
>>> iPhone[9]  # So the max index is 9, not 10
'6Plus'
>>> iPhone[10]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
IndexError: list index out of range
>>>

您需要这样做:

x = random.randint(0, 9)

，以便x始终在iPhone的有效索引范围内.

so that x will always be within the range of valid indexes for iPhone.

关于您的评论，您说您需要从列表中打印三个随机项目.因此，您可以执行以下操作:

Regarding your comments, you said that you need to print three random items from the list. So, you could do something like this:

import random

iPhone = 'Original 3G 3GS 4 4S 5 5C 5S 6 6Plus'.split()

z = 0
while z < 3:
    x = random.randint(0,9)
    print (iPhone[x])
    z += 1  # Remember to increment z so the while loop exits when it should

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